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Difference between revisions of "2006 AIME I Problems/Problem 1"

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== Problem ==
 
== Problem ==
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In [[convex polygon|convex]] [[hexagon]] <math>ABCDEF</math>, all six sides are congruent, <math>\angle A</math> and <math>\angle D</math> are [[right angle]]s, and <math>\angle B, \angle C, \angle E,</math> and <math>\angle F</math> are [[congruent]]. The area of the hexagonal region is <math>2116(\sqrt{2}+1).</math> Find <math>AB</math>.
 
  
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== Solution ==
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In [[quadrilateral]] <math>ABCD</math>, <math>\angle B</math> is a [[right angle]], [[diagonal]] <math>\overline{AC}</math> is [[perpendicular]] to <math>\overline{CD}</math>, <math>AB=18</math>, <math>BC=21</math>, and <math>CD=14</math>. Find the [[perimeter]] of <math>ABCD</math>.
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Let the side length be called <math>x</math>, so <math>x=AB=BC=CD=DE=EF=AF</math>.  
 
  
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[[Image:2006_I_AIME-1.png]]
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== Solution 1 ==
  
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The diagonal <math>BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}</math>. Then the areas of the triangles AFB and CDE in total are <math>\frac{x^2}{2}\cdot 2</math>,
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We construct the following diagram:
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and the area of the rectangle BCEF equals <math>x\cdot x\sqrt{2}=x^2\sqrt{2}</math>
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<asy>
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pathpen = black;
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pair C=(0,0),D=(0,-14),A=(-sqrt(765),0),B=IP(circle(C,21),circle(A,18));
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D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C);
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D(rightanglemark(A,C,D,40));
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D(rightanglemark(A,B,C,40));
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</asy><!--Asymptote by joml88-->
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Using the [[Pythagorean Theorem]], we get the following two equations:
 +
<cmath>AD^2 = AC^2 + CD^2</cmath>
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<cmath>AC^2 = AB^2 + BC^2</cmath>
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Substituting <math>AB^2 + BC^2</math> for <math>AC^2</math> gives us <math>AD^2 = AB^2 + BC^2 + CD^2</math>. Plugging in the given information, we get <math>AD^2 = 18^2 + 21^2 + 14^2 = 961 \implies AD = 31</math>, so the perimeter is <math>AB+BC+CD+AD = 18+21+14+31 = \boxed{084}</math>.
  
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Then we have to solve the equation
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== See Also ==
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<div style="text-align:center;">
 
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<math>2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2</math>.
 
  
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<math>2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)</math>
 
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<math>2116=x^2</math>
 
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<math>x=46</math></div>
 
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Therefore, <math>AB</math> is <math>046</math>.
 
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== See also ==
 
 
{{AIME box|year=2006|n=I|before=First Question|num-a=2}}
 
{{AIME box|year=2006|n=I|before=First Question|num-a=2}}
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{{MAA Notice}}
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[[Category:Intermediate Geometry Problems]]
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[[Category:Introductory Geometry Problems]]

Latest revision as of 16:10, 3 February 2025

Problem

In quadrilateral $ABCD$, $\angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$, $AB=18$, $BC=21$, and $CD=14$. Find the perimeter of $ABCD$.

Solution 1

We construct the following diagram: [asy] pathpen = black; pair C=(0,0),D=(0,-14),A=(-sqrt(765),0),B=IP(circle(C,21),circle(A,18)); D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C); D(rightanglemark(A,C,D,40)); D(rightanglemark(A,B,C,40)); [/asy] Using the Pythagorean Theorem, we get the following two equations: \[AD^2 = AC^2 + CD^2\] \[AC^2 = AB^2 + BC^2\] Substituting $AB^2 + BC^2$ for $AC^2$ gives us $AD^2 = AB^2 + BC^2 + CD^2$. Plugging in the given information, we get $AD^2 = 18^2 + 21^2 + 14^2 = 961 \implies AD = 31$, so the perimeter is $AB+BC+CD+AD = 18+21+14+31 = \boxed{084}$.

See Also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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